Summarize the peitinent information obtained by applying the graphing strategy and skelch the graph of $$ y=f(x) $$. $$ f(x)=\left(x^{2}+10\right)\left(100-x^{2}\right) $$

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**Domain and Symmetry**

- The function is given by  
  $$
  f(x) = (x^2+10)(100-x^2).
  $$
  Since it is a polynomial, its domain is all real numbers.

- Notice that  
  $$
  f(-x) = ((-x)^2+10)(100-(-x)^2) = (x^2+10)(100-x^2) = f(x),
  $$
  so $$ f(x) $$ is an even function (symmetric about the $$ y $$-axis).

---

**Intercepts**

- **$$ x $$-intercepts**: Set $$ f(x)=0 $$.  
  Since  
  $$
  f(x) = (x^2+10)(100-x^2),
  $$
  the factor $$ x^2+10 $$ is always positive for real $$ x $$. Thus, the zeros come from  
  $$
  100-x^2=0 \quad \Rightarrow \quad x^2=100 \quad \Rightarrow \quad x=\pm 10.
  $$

- **$$ y $$-intercept**: Compute $$ f(0) $$.  
  $$
  f(0) = (0^2+10)(100-0^2) = 10\times 100 = 1000.
  $$
  So the graph passes through $$(0, 1000)$$.

---

**Rewriting the Function**

Expanding gives  
$$
f(x) = (x^2+10)(100-x^2)= 100x^2 + 1000 - x^4 -10x^2 = -x^4 + 90x^2+1000.
$$

---

**Critical Points and Extrema**

1. **Finding the derivative:**

$$
   f(x) = -x^4+90x^2+1000 \quad \Rightarrow \quad f'(x) = -4x^3+180x.
   $$
   Factor the derivative:
   $$
   f'(x) = -4x(x^2-45)= -4x(x-\sqrt{45})(x+\sqrt{45}).
   $$
   The critical points are at:
   $$
   x=0 \quad \text{and} \quad x=\pm \sqrt{45} = \pm 3\sqrt{5}.
   $$

2. **Evaluating $$ f(x) $$ at the critical points:**

- At $$ x=0 $$:
     $$
     f(0)=1000.
     $$
   - At $$ x=\pm 3\sqrt{5} $$:
     
     First, compute  
     $$
     (3\sqrt{5})^2 = 9\times5=45 \quad \text{and} \quad (3\sqrt{5})^4 = 45^2=2025.
     $$
     Then,
     $$
     f(3\sqrt{5}) = -2025+90(45)+1000 = -2025+4050+1000=3025.
     $$
     By symmetry, $$ f(-3\sqrt{5}) = 3025 $$.

3. **Interpreting the Critical Points:**

- $$ x=\pm 3\sqrt{5} $$ are local maximum points with a maximum value of $$ 3025 $$.
   - $$ x=0 $$ is a local minimum within the interval between these local maxima, though note that the endpoints $$ x=\pm 10 $$ yield $$ f(\pm10)=0 $$ which are lower; however, those are not local extrema because the function is decreasing past $$ |x|=10 $$.

---

**End Behavior**

- The leading term in the expanded form $$ f(x)=-x^4+90x^2+1000 $$ is $$-x^4$$.
- Thus, as $$ x\to\pm\infty $$,
  $$
  f(x) \to -\infty.
  $$

---

**Graph Summary and Sketch Description**

- **Symmetry:** The graph is even (symmetric about the $$ y $$-axis).

- **Key Points:**
  - $$ x $$-intercepts: $$ \; x=-10 $$ and $$ x=10 $$ (where $$ f(x)=0 $$).
  - $$ y $$-intercept: $$(0,1000)$$.
  - Local maxima: $$\; x=\pm 3\sqrt{5} \approx \pm 6.708 $$ with $$ f(x)=3025 $$.
  - Local minimum (within $$|x|<10$$): at $$ (0, 1000) $$.
  - For $$|x|>10$$, the function becomes negative and falls toward $$-\infty$$.

- **Sketch Outline:**
  - Draw the $$ y $$-axis as the axis of symmetry.
  - Mark the zeros at $$(-10, 0)$$ and $$(10, 0)$$.
  - Plot the $$ y $$-intercept at $$(0, 1000)$$.
  - Plot the local maximum points at approximately $$(-6.71, 3025)$$ and $$(6.71, 3025)$$.
  - For $$ |x| > 10 $$, the curve drops below the $$ x $$-axis tapering off to $$-\infty$$.
  - Connect these points with a smooth curve, ensuring the symmetry and the proper end behavior as $$ x\to\pm\infty $$.

---

**Conclusion**

The graph of  
$$
f(x)=\left(x^{2}+10\right)\left(100-x^{2}\right)
$$
has the following pertinent features:

- Domain: All real numbers.
- Even symmetry about the $$ y $$-axis.
- $$ x $$-intercepts at $$ x=\pm 10 $$.
- $$ y $$-intercept at $$(0, 1000)$$.
- Local maximum points at $$ x=\pm 3\sqrt{5} $$ with $$ f(x)=3025 $$.
- A local minimum at $$ x=0 $$ with $$ f(x)=1000 $$ (on the interval $$[-10,10]$$).
- As $$ x\to\pm\infty $$, $$ f(x)\to -\infty $$.

This information guides the sketch of the graph, ensuring accuracy in the locations of intercepts, extrema, symmetry, and end behavior.
The graph of $$ f(x) = (x^2 + 10)(100 - x^2) $$ is symmetric about the y-axis. It has x-intercepts at $$ x = \pm 10 $$ and a y-intercept at $$ (0, 1000) $$. There are local maximum points at $$ x = \pm 3\sqrt{5} $$ with a value of 3025. The function has a local minimum at $$ x = 0 $$ with a value of 1000. As $$ x $$ approaches positive or negative infinity, $$ f(x) $$ decreases without bound.

Summarize the peitinent information obtained by applying the graphing strategy and skelch the graph of \( y=f(x) \). \( f(x)=\left(x^{2}+10\right)\left(100-x^{2}\right) \)

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